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pychu [463]
3 years ago
8

Consider the situation||: A child pulls a sled by a rope across the lawn at a constant speed. Of the forces listed, identify whi

ch act upon the sled.
Normal, Gravity, Applied, Friction, Tension, & Air Resistance
Physics
1 answer:
Mice21 [21]3 years ago
6 0

Answer:

Gravitational

Tension

Normal

Friction.

Explanation:

The forces acting on the sled are:

Tension: the tension from the rope, this is the force that "moves" the sled.

Friction: kinetic friction between the sled and the ground as the sled moves.

There are another two forces that also act on the sled, but that "has no effect"

Gravitational force: This force pulls the sled down, against the floor.

Normal force: This force "opposes" to the gravitational one, so they cancel each other.

These two forces cancel each other, so they have no direct impact on the movement of the sled. BUT, the friction force depends on the weight of the moving object, and the weight of the moving object depends on the gravitational force, so we need gravitational force in order to have friction force.

Then we can conclude that the forces acting on the sled are:

Gravitational

Tension

Normal

Friction.

You might be interested in
A banana peel has lots of friction.<br> True or False
olga_2 [115]

Answer:

False

Explanation:

I learned it the hard way trust me T^T

3 0
3 years ago
suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area vapour already present in su
gulaghasi [49]
That's two different things it depends on:

-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.

Here's what I have in mind for an experiment to show those two dependencies:

-- a closed box with a wall down the middle, separating it into two closed sections;

-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.

-- a tiny fan that blows air through a tube into the hole in one outer wall.

<u>Experiment A:</u>

-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================

<u>Experiment B:</u>

-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section.  Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
6 0
3 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
Driving Zone 7 corresponds with
stepan [7]
Your driving zone refers to the areas of space around your car, it refers to all the area around your car as far as your eyes can see. 
Each car has seven zones numbered from 1 to 7. Driving zone 7 corresponds with THE SPACE YOUR VEHICLE IS OCCUPYING. The other zones are as follows:
zone 1 = area directly infront of your car
zone 2 = your left lane
zone 3 = your right lane
zone 4 = left rear of your car
zone 5 = right rear of your car 
zone 6 = area directly behind your car.
<span />
3 0
3 years ago
The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i
spayn [35]

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

7 0
3 years ago
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